Question 1180838
.
A metallurgist has one alloy containing 22% copper and another containing 53% copper. 
How many pounds of each alloy must he use to make 51 pounds of a third alloy containing 30% copper? 
(Round to two decimal places if necessary.)
~~~~~~~~~~~~~


<pre>
Let x be the mass of the 53% copper alloy (in pounds).

Then the mass of the 22% copper alloy is (51-x) pounds.


Now you write equation, saying that the sum of amounts of pure copper in ingredients
is the same as that in the final alloy


    0.53x + 0.22*(51-x) = 0.3*51.


From the equation


    x = {{{(0.3*51-0.22*51)/(0.53 - 0.22)}}} = 13.16  pounds (rounded).


<U>ANSWER</U>.  13.16 pounds of the 53% copper alloy and the rest,  (51-13.16) = 37.84 pounds of the 22% copper alloy.


<U>CHECK</U>.   I will check the concentration of the final alloy  {{{(0.53*13.16 + 0.22*37.84)/51}}} = 0.30, or 30%.
</pre>

Solved.